3.325 \(\int e^{\coth ^{-1}(x)} \sqrt {1-x} x \, dx\)

Optimal. Leaf size=71 \[ \frac {2 \left (\frac {1}{x}+1\right )^{3/2} \sqrt {1-x} x^2}{5 \sqrt {1-\frac {1}{x}}}-\frac {4 \left (\frac {1}{x}+1\right )^{3/2} \sqrt {1-x} x}{15 \sqrt {1-\frac {1}{x}}} \]

[Out]

-4/15*(1/x+1)^(3/2)*x*(1-x)^(1/2)/(1-1/x)^(1/2)+2/5*(1/x+1)^(3/2)*x^2*(1-x)^(1/2)/(1-1/x)^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {6176, 6181, 45, 37} \[ \frac {2 \left (\frac {1}{x}+1\right )^{3/2} \sqrt {1-x} x^2}{5 \sqrt {1-\frac {1}{x}}}-\frac {4 \left (\frac {1}{x}+1\right )^{3/2} \sqrt {1-x} x}{15 \sqrt {1-\frac {1}{x}}} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCoth[x]*Sqrt[1 - x]*x,x]

[Out]

(-4*(1 + x^(-1))^(3/2)*Sqrt[1 - x]*x)/(15*Sqrt[1 - x^(-1)]) + (2*(1 + x^(-1))^(3/2)*Sqrt[1 - x]*x^2)/(5*Sqrt[1
- x^(-1)])

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 6176

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_))^(p_), x_Symbol] :> Dist[(c + d*x)^p/(x^p*(1 + c/(d
*x))^p), Int[u*x^p*(1 + c/(d*x))^p*E^(n*ArcCoth[a*x]), x], x] /; FreeQ[{a, c, d, n, p}, x] && EqQ[a^2*c^2 - d^
2, 0] &&  !IntegerQ[n/2] &&  !IntegerQ[p]

Rule 6181

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_), x_Symbol] :> -Dist[c^p*x^m*(1/x)^m, Sub
st[Int[((1 + (d*x)/c)^p*(1 + x/a)^(n/2))/(x^(m + 2)*(1 - x/a)^(n/2)), x], x, 1/x], x] /; FreeQ[{a, c, d, m, n,
p}, x] && EqQ[c^2 - a^2*d^2, 0] &&  !IntegerQ[n/2] && (IntegerQ[p] || GtQ[c, 0]) &&  !IntegerQ[m]

Rubi steps

\begin {align*} \int e^{\coth ^{-1}(x)} \sqrt {1-x} x \, dx &=\frac {\sqrt {1-x} \int e^{\coth ^{-1}(x)} \sqrt {1-\frac {1}{x}} x^{3/2} \, dx}{\sqrt {1-\frac {1}{x}} \sqrt {x}}\\ &=-\frac {\left (\sqrt {1-x} \sqrt {\frac {1}{x}}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1+x}}{x^{7/2}} \, dx,x,\frac {1}{x}\right )}{\sqrt {1-\frac {1}{x}}}\\ &=\frac {2 \left (1+\frac {1}{x}\right )^{3/2} \sqrt {1-x} x^2}{5 \sqrt {1-\frac {1}{x}}}+\frac {\left (2 \sqrt {1-x} \sqrt {\frac {1}{x}}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1+x}}{x^{5/2}} \, dx,x,\frac {1}{x}\right )}{5 \sqrt {1-\frac {1}{x}}}\\ &=-\frac {4 \left (1+\frac {1}{x}\right )^{3/2} \sqrt {1-x} x}{15 \sqrt {1-\frac {1}{x}}}+\frac {2 \left (1+\frac {1}{x}\right )^{3/2} \sqrt {1-x} x^2}{5 \sqrt {1-\frac {1}{x}}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 41, normalized size = 0.58 \[ \frac {2 \sqrt {\frac {1}{x}+1} \sqrt {1-x} (x+1) (3 x-2)}{15 \sqrt {\frac {x-1}{x}}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^ArcCoth[x]*Sqrt[1 - x]*x,x]

[Out]

(2*Sqrt[1 + x^(-1)]*Sqrt[1 - x]*(1 + x)*(-2 + 3*x))/(15*Sqrt[(-1 + x)/x])

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fricas [A]  time = 0.63, size = 40, normalized size = 0.56 \[ \frac {2 \, {\left (3 \, x^{3} + 4 \, x^{2} - x - 2\right )} \sqrt {-x + 1} \sqrt {\frac {x - 1}{x + 1}}}{15 \, {\left (x - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((-1+x)/(1+x))^(1/2)*x*(1-x)^(1/2),x, algorithm="fricas")

[Out]

2/15*(3*x^3 + 4*x^2 - x - 2)*sqrt(-x + 1)*sqrt((x - 1)/(x + 1))/(x - 1)

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giac [C]  time = 0.18, size = 44, normalized size = 0.62 \[ \frac {1}{15} \, {\left (-4 i \, \sqrt {2} - \frac {2 \, {\left (3 \, {\left (x + 1\right )}^{2} \sqrt {-x - 1} + 5 \, {\left (-x - 1\right )}^{\frac {3}{2}}\right )}}{\mathrm {sgn}\left (-x - 1\right )}\right )} \mathrm {sgn}\relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((-1+x)/(1+x))^(1/2)*x*(1-x)^(1/2),x, algorithm="giac")

[Out]

1/15*(-4*I*sqrt(2) - 2*(3*(x + 1)^2*sqrt(-x - 1) + 5*(-x - 1)^(3/2))/sgn(-x - 1))*sgn(x)

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maple [A]  time = 0.03, size = 29, normalized size = 0.41 \[ \frac {2 \left (1+x \right ) \left (3 x -2\right ) \sqrt {1-x}}{15 \sqrt {\frac {-1+x}{1+x}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((-1+x)/(1+x))^(1/2)*x*(1-x)^(1/2),x)

[Out]

2/15*(1+x)*(3*x-2)*(1-x)^(1/2)/((-1+x)/(1+x))^(1/2)

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maxima [C]  time = 1.28, size = 17, normalized size = 0.24 \[ \frac {1}{15} \, {\left (6 i \, x^{2} + 2 i \, x - 4 i\right )} \sqrt {x + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((-1+x)/(1+x))^(1/2)*x*(1-x)^(1/2),x, algorithm="maxima")

[Out]

1/15*(6*I*x^2 + 2*I*x - 4*I)*sqrt(x + 1)

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mupad [B]  time = 1.27, size = 30, normalized size = 0.42 \[ -\frac {2\,\left (3\,x-2\right )\,\sqrt {\frac {x-1}{x+1}}\,{\left (x+1\right )}^2}{15\,\sqrt {1-x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(1 - x)^(1/2))/((x - 1)/(x + 1))^(1/2),x)

[Out]

-(2*(3*x - 2)*((x - 1)/(x + 1))^(1/2)*(x + 1)^2)/(15*(1 - x)^(1/2))

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sympy [C]  time = 16.15, size = 46, normalized size = 0.65 \[ - \frac {14 i x}{15 \sqrt {\frac {1}{x + 1}}} - \frac {2 i \left (1 - x\right )^{2}}{5 \sqrt {\frac {1}{x + 1}}} + \frac {2 i}{3 \sqrt {\frac {1}{x + 1}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((-1+x)/(1+x))**(1/2)*x*(1-x)**(1/2),x)

[Out]

-14*I*x/(15*sqrt(1/(x + 1))) - 2*I*(1 - x)**2/(5*sqrt(1/(x + 1))) + 2*I/(3*sqrt(1/(x + 1)))

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